nAl =\(\dfrac{5.4}{27}\)=0.2 (mol) đổi 200ml = 0,2l

nH2SO4 = Cm.V =1,35.0,2=0,27(MOL)

2Al + 3H2SO4\(\overrightarrow{ }\)Al2(SO4)3 + 3H2

pt; 2 ; 3 : 1 : 3

đb; 0.18 : 0.27 : 0.09 : 0.27 (mol)

so sánh nAl =\(\dfrac{0.2}{2}\)>nH2SO4 =\(\dfrac{0.27}{3}\)

a, nAl dư = 0.2-0.18=0.02(mol)

m Al dư = 0,02.27=0.54(g)

b, V\(H\)\(2\)=0,27.22,4 = 6,048(l)

c, dd tạo thành sau pư là Al2(SO4)3

Cm Al2(SO4)3 = \(\dfrac{n}{V}\)=\(\dfrac{0.09}{0.2}\)=0.45M

THE END